Problem: $A=\left[\begin{array}{rr}20 & -12 & 23 & 13 \\-8 & 12 & -6 & 6 \\0 &1 &38 & 1 \\-9 &5 &14 & 4\end{array}\right]$ $A_{3,4}=$
Background An $m\times n$ matrix has $m$ rows and $n$ columns. $A=\left[\begin{array}{rr}A_{1,1} & \cdots & A_{1,n} \\\\\vdots \ & \ddots & \vdots \\\\A_{m,1} &\cdots &A_{m,n}\end{array}\right]$ Therefore, the entry $A_{{c},{d}}$ is located on row ${c}$ and column ${d}$. Finding $A_{3,4}$ $A_{{3},{4}}$ is located on row ${3}$ of $A$ : $\left[\begin{array}{rr}20 & -12 & 23 & 13 \\-8 & 12 & -6 & 6 \\ {0} & {1} & {38} & {1} \\-9 &5 &14 & 4\end{array}\right]$ $A_{{3},{4}}$ is also located on column ${4}$ of $A$. $\left[\begin{array}{rr}20 & -12 & 23 & {13} \\-8 & 12 & -6 & {6} \\ {0} & {1} & {38} & {\text{1}} \\-9 &5 &14 & {4}\end{array}\right]$ Therefore, $A_{{3},{4}}={1}$. Summary $A_{3,4}=1$